Your student government association decided to do a fundraiser to raise
money for a field trip. they decide to sell t-shirts and sweatshirts. the
profit for each t-shirt is $10 and the profit for each sweatshirt is $15.
they want to sell 50 items at most. compared to t-shirts, they want to sell
at least half as many sweatshirts, with a profit of at least $500.
after graphing what are the boundaries of the feasible region (i. e. the points that define
the region of solutions)? (hint: there should be four points)
are there any non-viable solutions within the feasible region
how many t-shirts and sweatshirts should they sell to maximize their profit
what is the maximum profit they can make?

Step-by-step explanation:

If x is the number of t-shirts, and y is the number of sweatshirts:

10x + 15y ≥ 500

x + y ≤ 50

y ≥ x/2

And since x and y can't be negative:

x ≥ 0

y ≥ 0

Graph:

desmos.com/calculator/mw6dsei6jm

The boundaries of the feasible region are:

(0, 50), (0, 33.3), (28.6, 14.3), and (33.3, 16.7)

Since x and y must be integers, any non-integer solutions are not viable.

The maximum profit will be when the most product is sold.  Since sweatshirts are more profitable, we want to sell as much of that as we can.  

So they should sell 0 t-shirts and 50 sweatshirts for a maximum profit of $750.

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1